WBJEE · Maths · Circle
If one of the diameters of the curve \(x^{2}+y^{2}-4 x-6 y+9=0\) is a chord of a circle with centre \((1,1),\) the radius of this circle is
- A 3
- B 2
- C \(\sqrt{2}\)
- D 1
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
Given circle \(x^{2}+y^{2}-4 x-6 y+9=0\) \(\therefore\) Centre \(=\left(-\frac{1}{2} \times(-4),-\frac{1}{2} \times(-6)\right)\) \(=(2.3)\) \(\begin{aligned} \text { Radius } &=\sqrt{(-2)^{2}+(-3)^{2}-9} \\ &=\sqrt{4+9-9} \\ &=2 \end{aligned}\)…
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